The integral $\int \cos \left(\log _{e} x\right) d x$ is equal to: (where $C$ is a constant of integration)

(C) $\frac{x}{2}\left[\cos \left(\log _{e} x\right)+\sin \left(\log _{e} x\right)\right]+C$

(A) $\frac{x}{2}\left[\sin \left(\log _{e} x\right)-\cos \left(\log _{e} x\right)\right]+C$

(B) $x\left[\cos \left(\log _{e} x\right)+\sin \left(\log _{e} x\right)\right]+C$

(D) $x\left[\cos \left(\log _{e} x\right)-\sin \left(\log _{e} x\right)\right]+C$

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Indefinite Integrals

The integral

\int cos (lo g_{e} x) d x

is equal to: (where

C

is a constant of integration)

Answer: (C)

For

I = \int cos (ln x) d x

, integrate by parts with

u = cos (ln x)

and

d v = d x

I = x cos (ln x) + \int sin (ln x) d x

. Call the new integral

J = \int sin (ln x) d x

and integrate it by parts too:

J = x sin (ln x) - \int cos (ln x) d x = x sin (ln x) - I

. Substituting back,

I = x cos (ln x) + x sin (ln x) - I

, so

2 I = x [cos (ln x) + sin (ln x)]

and

I = \frac{x}{2} [cos (ln x) + sin (ln x)] + C

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option